Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x + 5} = \dfrac{-2x + 79}{x + 5}$
Multiply both sides by $x + 5$ $ \dfrac{x^2 - 1}{x + 5} (x + 5) = \dfrac{-2x + 79}{x + 5} (x + 5)$ $ x^2 - 1 = -2x + 79$ Subtract $-2x + 79$ from both sides: $ x^2 - 1 - (-2x + 79) = -2x + 79 - (-2x + 79)$ $ x^2 - 1 + 2x - 79 = 0$ $ x^2 - 80 + 2x = 0$ Factor the expression: $ (x + 10)(x - 8) = 0$ Therefore $x = -10$ or $x = 8$ The original expression is defined at $x = -10$ and $x = 8$, so there are no extraneous solutions.